From 170c10f952ee8f5efe837fe806df45332a086226 Mon Sep 17 00:00:00 2001 From: yanglbme Date: Thu, 23 Mar 2023 09:37:56 +0800 Subject: [PATCH] feat: add solutions to lc problem: No.0629 No.0629.K Inverse Pairs Array --- .../0629.K Inverse Pairs Array/README.md | 201 +++++++----------- .../0629.K Inverse Pairs Array/README_EN.md | 176 ++++++--------- .../0629.K Inverse Pairs Array/Solution.cpp | 23 +- .../0629.K Inverse Pairs Array/Solution.go | 22 +- .../0629.K Inverse Pairs Array/Solution.java | 29 ++- .../0629.K Inverse Pairs Array/Solution.py | 16 +- .../0629.K Inverse Pairs Array/Solution.ts | 16 ++ .../1630.Arithmetic Subarrays/README.md | 2 +- 8 files changed, 203 insertions(+), 282 deletions(-) create mode 100644 solution/0600-0699/0629.K Inverse Pairs Array/Solution.ts diff --git a/solution/0600-0699/0629.K Inverse Pairs Array/README.md b/solution/0600-0699/0629.K Inverse Pairs Array/README.md index 9374a16f2e768..80b12efd75b2a 100644 --- a/solution/0600-0699/0629.K Inverse Pairs Array/README.md +++ b/solution/0600-0699/0629.K Inverse Pairs Array/README.md @@ -40,18 +40,25 @@ -动态规划，我们规定 `dp[i][j]` 表示 `i` 个数字恰好拥有 `j` 个逆序对的不同数组的个数，最终答案为 `dp[n][k]` +**方法一：动态规划 + 前缀和** -思考如何得到 `dp[i][j]`，假设 `i - 1` 个数字已经确定，现在插入 `i` 一共有 `i` 种情况： +我们定义 $f[i][j]$ 表示数组长度为 $i$，逆序对数为 $j$ 的数组个数。初始时 $f[0][0] = 1$，其余 $f[i][j] = 0$。 -- 放在第一个，由于 `i` 比之前的任何数都大，所以会产生 `i - 1` 个逆序对，为了凑够 `j` 个逆序对，之前确定的数需要产生 `j - (i - 1)` 个逆序对 -- 放在第二个，产生 `i - 2` 个逆序对，为了凑够 `j` 个逆序对，之前确定的数需要产生 `j - (i - 2)` 个逆序对 -- 放在第三个......同理 -- 放在最后一个，产生 `0` 个逆序对，之前确认的数需要产生 `j` 个逆序对 +接下来我们考虑如何得到 $f[i][j]$。 -可得状态转移公式：`dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j]` +假设前 $i-1$ 个数已经确定，现在要插入数字 $i$，我们讨论 $i$ 插入到每个位置的情况： -看到这种累加，很容易想到需要用前缀和进行优化。最终 `dp[i][]` 只依赖前缀和数组，甚至连 `dp[i - 1][]` 都不需要，所以可以进一步用一维数组优化空间 +- 如果 $i$ 插入到第 $1$ 个位置，那么逆序对增加了 $i-1$ 个，所以 $f[i][j]+=f[i-1][j-(i-1)]$。 +- 如果 $i$ 插入到第 $2$ 个位置，那么逆序对增加了 $i-2$ 个，所以 $f[i][j]+=f[i-1][j-(i-2)]$。 +- ... +- 如果 $i$ 插入到第 $i-1$ 个位置，那么逆序对增加了 $1$ 个，所以 $f[i][j]+=f[i-1][j-1]$。 +- 如果 $i$ 插入到第 $i$ 个位置，那么逆序对不变，所以 $f[i][j]+=f[i-1][j]$。 + +所以 $f[i][j]=\sum_{k=1}^{i}f[i-1][j-(i-k)]$。 + +我们注意到 $f[i][j]$ 的计算实际上涉及到前缀和，因此，我们可以使用前缀和优化计算过程。并且，由于 $f[i][j]$ 只与 $f[i-1][j]$ 有关，因此我们可以用一个一维数组来优化空间复杂度。 + +时间复杂度 $O(n \times k)$，空间复杂度 $O(k)$。其中 $n$ 和 $k$ 分别为数组长度和逆序对数。 @@ -62,59 +69,15 @@ ```python class Solution: def kInversePairs(self, n: int, k: int) -> int: - mod = 1000000007 - dp, pre = [0] * (k + 1), [0] * (k + 2) + mod = 10**9 + 7 + f = [1] + [0] * k + s = [0] * (k + 2) for i in range(1, n + 1): - dp[0] = 1 - - # dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] for j in range(1, k + 1): - dp[j] = (pre[j + 1] - pre[max(0, j - i + 1)] + mod) % mod - + f[j] = (s[j + 1] - s[max(0, j - (i - 1))]) % mod for j in range(1, k + 2): - pre[j] = (pre[j - 1] + dp[j - 1]) % mod - - return dp[k] -``` - -`dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] + dp[i - 1][j - 2] + ... + dp[i - 1][j - (i - 1)]` ① - -`dp[i][j - 1] = dp[i - 1][j - 1] + dp[i - 1][j - 2] + ... + dp[i - 1][j - (i - 1)] + dp[i - 1][j - i]` ② - -① - ②，得 `dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - i]` - -```python -class Solution: - def kInversePairs(self, n: int, k: int) -> int: - N, MOD = 1010, int(1e9) + 7 - dp = [[0] * N for _ in range(N)] - dp[1][0] = 1 - for i in range(2, n + 1): - dp[i][0] = 1 - for j in range(1, k + 1): - dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - if j >= i: - dp[i][j] -= dp[i - 1][j - i] - dp[i][j] %= MOD - return dp[n][k] -``` - -空间优化： - -```python -class Solution: - def kInversePairs(self, n: int, k: int) -> int: - N, MOD = 1010, int(1e9) + 7 - dp = [0] * N - dp[0] = 1 - for i in range(2, n + 1): - t = dp.copy() - for j in range(1, k + 1): - dp[j] = t[j] + dp[j - 1] - if j >= i: - dp[j] -= t[j - i] - dp[j] %= MOD - return dp[k] + s[j] = (s[j - 1] + f[j - 1]) % mod + return f[k] ``` ### **Java** @@ -123,70 +86,72 @@ class Solution: ```java class Solution { - - private static final int MOD = 1000000007; - public int kInversePairs(int n, int k) { - int[] dp = new int[k + 1]; - int[] pre = new int[k + 2]; - for (int i = 1; i <= n; i++) { - dp[0] = 1; - - // dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] - for (int j = 1; j <= k; j++) { - dp[j] = (pre[j + 1] - pre[Math.max(0, j - i + 1)] + MOD) % MOD; + final int mod = (int) 1e9 + 7; + int[] f = new int[k + 1]; + int[] s = new int[k + 2]; + f[0] = 1; + Arrays.fill(s, 1); + s[0] = 0; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= k; ++j) { + f[j] = (s[j + 1] - s[Math.max(0, j - (i - 1))] + mod) % mod; } - - for (int j = 1; j <= k + 1; j++) { - pre[j] = (pre[j - 1] + dp[j - 1]) % MOD; + for (int j = 1; j <= k + 1; ++j) { + s[j] = (s[j - 1] + f[j - 1]) % mod; } } - return dp[k]; + return f[k]; } } ``` -```java +### **C++** + +```cpp class Solution { - public int kInversePairs(int n, int k) { - int N = 1010, MOD = (int) (1e9 + 7); - int[][] dp = new int[N][N]; - dp[1][0] = 1; - for (int i = 2; i < n + 1; ++i) { - dp[i][0] = 1; - for (int j = 1; j < k + 1; ++j) { - dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MOD; - if (j >= i) { - dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + MOD) % MOD; - } +public: + int kInversePairs(int n, int k) { + int f[k + 1]; + int s[k + 2]; + memset(f, 0, sizeof(f)); + f[0] = 1; + fill(s, s + k + 2, 1); + s[0] = 0; + const int mod = 1e9 + 7; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= k; ++j) { + f[j] = (s[j + 1] - s[max(0, j - (i - 1))] + mod) % mod; + } + for (int j = 1; j <= k + 1; ++j) { + s[j] = (s[j - 1] + f[j - 1]) % mod; } } - return dp[n][k]; + return f[k]; } -} +}; ``` ### **Go** ```go -const mod int = 1e9 + 7 - func kInversePairs(n int, k int) int { - dp := make([]int, k+1) - pre := make([]int, k+2) + f := make([]int, k+1) + s := make([]int, k+2) + f[0] = 1 + for i, x := range f { + s[i+1] = s[i] + x + } + const mod = 1e9 + 7 for i := 1; i <= n; i++ { - dp[0] = 1 - - // dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] for j := 1; j <= k; j++ { - dp[j] = (pre[j+1] - pre[max(0, j-i+1)] + mod) % mod + f[j] = (s[j+1] - s[max(0, j-(i-1))] + mod) % mod } - for j := 1; j <= k+1; j++ { - pre[j] = (pre[j-1] + dp[j-1]) % mod + s[j] = (s[j-1] + f[j-1]) % mod } } - return dp[k] + return f[k] } func max(a, b int) int { @@ -197,31 +162,25 @@ func max(a, b int) int { } ``` -### **C++** - -```cpp -class Solution { -private: - static constexpr int MOD = 1e9 + 7; - -public: - int kInversePairs(int n, int k) { - vector dp(k + 1), pre(k + 2, 0); - for (int i = 1; i <= n; ++i) { - dp[0] = 1; - - // dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] - for (int j = 1; j <= k; ++j) { - dp[j] = (pre[j + 1] - pre[max(0, j - i + 1)] + MOD) % MOD; - } - - for (int j = 1; j <= k + 1; ++j) { - pre[j] = (pre[j - 1] + dp[j - 1]) % MOD; - } +### **TypeScript** + +```ts +function kInversePairs(n: number, k: number): number { + const f: number[] = new Array(k + 1).fill(0); + f[0] = 1; + const s: number[] = new Array(k + 2).fill(1); + s[0] = 0; + const mod: number = 1e9 + 7; + for (let i = 1; i <= n; ++i) { + for (let j = 1; j <= k; ++j) { + f[j] = (s[j + 1] - s[Math.max(0, j - (i - 1))] + mod) % mod; + } + for (let j = 1; j <= k + 1; ++j) { + s[j] = (s[j - 1] + f[j - 1]) % mod; } - return dp[k]; } -}; + return f[k]; +} ``` ### **...** diff --git a/solution/0600-0699/0629.K Inverse Pairs Array/README_EN.md b/solution/0600-0699/0629.K Inverse Pairs Array/README_EN.md index c97ac3ea6b211..deab8a8c54df9 100644 --- a/solution/0600-0699/0629.K Inverse Pairs Array/README_EN.md +++ b/solution/0600-0699/0629.K Inverse Pairs Array/README_EN.md @@ -42,127 +42,87 @@ ```python class Solution: def kInversePairs(self, n: int, k: int) -> int: - mod = 1000000007 - dp, pre = [0] * (k + 1), [0] * (k + 2) + mod = 10**9 + 7 + f = [1] + [0] * k + s = [0] * (k + 2) for i in range(1, n + 1): - dp[0] = 1 - - # dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] for j in range(1, k + 1): - dp[j] = (pre[j + 1] - pre[max(0, j - i + 1)] + mod) % mod - + f[j] = (s[j + 1] - s[max(0, j - (i - 1))]) % mod for j in range(1, k + 2): - pre[j] = (pre[j - 1] + dp[j - 1]) % mod - - return dp[k] -``` - -`dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] + dp[i - 1][j - 2] + ... + dp[i - 1][j - (i - 1)]` ① - -`dp[i][j - 1] = dp[i - 1][j - 1] + dp[i - 1][j - 2] + ... + dp[i - 1][j - (i - 1)] + dp[i - 1][j - i]` ② - -① - ②, `dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - i]` - -```python -class Solution: - def kInversePairs(self, n: int, k: int) -> int: - N, MOD = 1010, int(1e9) + 7 - dp = [[0] * N for _ in range(N)] - dp[1][0] = 1 - for i in range(2, n + 1): - dp[i][0] = 1 - for j in range(1, k + 1): - dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - if j >= i: - dp[i][j] -= dp[i - 1][j - i] - dp[i][j] %= MOD - return dp[n][k] -``` - -```python -class Solution: - def kInversePairs(self, n: int, k: int) -> int: - N, MOD = 1010, int(1e9) + 7 - dp = [0] * N - dp[0] = 1 - for i in range(2, n + 1): - t = dp.copy() - for j in range(1, k + 1): - dp[j] = t[j] + dp[j - 1] - if j >= i: - dp[j] -= t[j - i] - dp[j] %= MOD - return dp[k] + s[j] = (s[j - 1] + f[j - 1]) % mod + return f[k] ``` ### **Java** ```java class Solution { - - private static final int MOD = 1000000007; - public int kInversePairs(int n, int k) { - int[] dp = new int[k + 1]; - int[] pre = new int[k + 2]; - for (int i = 1; i <= n; i++) { - dp[0] = 1; - - // dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] - for (int j = 1; j <= k; j++) { - dp[j] = (pre[j + 1] - pre[Math.max(0, j - i + 1)] + MOD) % MOD; + final int mod = (int) 1e9 + 7; + int[] f = new int[k + 1]; + int[] s = new int[k + 2]; + f[0] = 1; + Arrays.fill(s, 1); + s[0] = 0; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= k; ++j) { + f[j] = (s[j + 1] - s[Math.max(0, j - (i - 1))] + mod) % mod; } - - for (int j = 1; j <= k + 1; j++) { - pre[j] = (pre[j - 1] + dp[j - 1]) % MOD; + for (int j = 1; j <= k + 1; ++j) { + s[j] = (s[j - 1] + f[j - 1]) % mod; } } - return dp[k]; + return f[k]; } } ``` -```java +### **C++** + +```cpp class Solution { - public int kInversePairs(int n, int k) { - int N = 1010, MOD = (int) (1e9 + 7); - int[][] dp = new int[N][N]; - dp[1][0] = 1; - for (int i = 2; i < n + 1; ++i) { - dp[i][0] = 1; - for (int j = 1; j < k + 1; ++j) { - dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MOD; - if (j >= i) { - dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + MOD) % MOD; - } +public: + int kInversePairs(int n, int k) { + int f[k + 1]; + int s[k + 2]; + memset(f, 0, sizeof(f)); + f[0] = 1; + fill(s, s + k + 2, 1); + s[0] = 0; + const int mod = 1e9 + 7; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= k; ++j) { + f[j] = (s[j + 1] - s[max(0, j - (i - 1))] + mod) % mod; + } + for (int j = 1; j <= k + 1; ++j) { + s[j] = (s[j - 1] + f[j - 1]) % mod; } } - return dp[n][k]; + return f[k]; } -} +}; ``` ### **Go** ```go -const mod int = 1e9 + 7 - func kInversePairs(n int, k int) int { - dp := make([]int, k+1) - pre := make([]int, k+2) + f := make([]int, k+1) + s := make([]int, k+2) + f[0] = 1 + for i, x := range f { + s[i+1] = s[i] + x + } + const mod = 1e9 + 7 for i := 1; i <= n; i++ { - dp[0] = 1 - - // dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] for j := 1; j <= k; j++ { - dp[j] = (pre[j+1] - pre[max(0, j-i+1)] + mod) % mod + f[j] = (s[j+1] - s[max(0, j-(i-1))] + mod) % mod } - for j := 1; j <= k+1; j++ { - pre[j] = (pre[j-1] + dp[j-1]) % mod + s[j] = (s[j-1] + f[j-1]) % mod } } - return dp[k] + return f[k] } func max(a, b int) int { @@ -173,31 +133,25 @@ func max(a, b int) int { } ``` -### **C++** - -```cpp -class Solution { -private: - static constexpr int MOD = 1e9 + 7; - -public: - int kInversePairs(int n, int k) { - vector dp(k + 1), pre(k + 2, 0); - for (int i = 1; i <= n; ++i) { - dp[0] = 1; - - // dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] - for (int j = 1; j <= k; ++j) { - dp[j] = (pre[j + 1] - pre[max(0, j - i + 1)] + MOD) % MOD; - } - - for (int j = 1; j <= k + 1; ++j) { - pre[j] = (pre[j - 1] + dp[j - 1]) % MOD; - } +### **TypeScript** + +```ts +function kInversePairs(n: number, k: number): number { + const f: number[] = new Array(k + 1).fill(0); + f[0] = 1; + const s: number[] = new Array(k + 2).fill(1); + s[0] = 0; + const mod: number = 1e9 + 7; + for (let i = 1; i <= n; ++i) { + for (let j = 1; j <= k; ++j) { + f[j] = (s[j + 1] - s[Math.max(0, j - (i - 1))] + mod) % mod; + } + for (let j = 1; j <= k + 1; ++j) { + s[j] = (s[j - 1] + f[j - 1]) % mod; } - return dp[k]; } -}; + return f[k]; +} ``` ### **...** diff --git a/solution/0600-0699/0629.K Inverse Pairs Array/Solution.cpp b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.cpp index 40b1e73cfca7c..2553de10e8982 100644 --- a/solution/0600-0699/0629.K Inverse Pairs Array/Solution.cpp +++ b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.cpp @@ -1,22 +1,21 @@ class Solution { -private: - static constexpr int MOD = 1e9 + 7; - public: int kInversePairs(int n, int k) { - vector dp(k + 1), pre(k + 2, 0); + int f[k + 1]; + int s[k + 2]; + memset(f, 0, sizeof(f)); + f[0] = 1; + fill(s, s + k + 2, 1); + s[0] = 0; + const int mod = 1e9 + 7; for (int i = 1; i <= n; ++i) { - dp[0] = 1; - - // dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] for (int j = 1; j <= k; ++j) { - dp[j] = (pre[j + 1] - pre[max(0, j - i + 1)] + MOD) % MOD; + f[j] = (s[j + 1] - s[max(0, j - (i - 1))] + mod) % mod; } - for (int j = 1; j <= k + 1; ++j) { - pre[j] = (pre[j - 1] + dp[j - 1]) % MOD; + s[j] = (s[j - 1] + f[j - 1]) % mod; } } - return dp[k]; + return f[k]; } -}; +}; \ No newline at end of file diff --git a/solution/0600-0699/0629.K Inverse Pairs Array/Solution.go b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.go index 6af80a6123b1d..fbd9788feff38 100644 --- a/solution/0600-0699/0629.K Inverse Pairs Array/Solution.go +++ b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.go @@ -1,21 +1,20 @@ -const mod int = 1e9 + 7 - func kInversePairs(n int, k int) int { - dp := make([]int, k+1) - pre := make([]int, k+2) + f := make([]int, k+1) + s := make([]int, k+2) + f[0] = 1 + for i, x := range f { + s[i+1] = s[i] + x + } + const mod = 1e9 + 7 for i := 1; i <= n; i++ { - dp[0] = 1 - - // dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] for j := 1; j <= k; j++ { - dp[j] = (pre[j+1] - pre[max(0, j-i+1)] + mod) % mod + f[j] = (s[j+1] - s[max(0, j-(i-1))] + mod) % mod } - for j := 1; j <= k+1; j++ { - pre[j] = (pre[j-1] + dp[j-1]) % mod + s[j] = (s[j-1] + f[j-1]) % mod } } - return dp[k] + return f[k] } func max(a, b int) int { @@ -24,3 +23,4 @@ func max(a, b int) int { } return b } + diff --git a/solution/0600-0699/0629.K Inverse Pairs Array/Solution.java b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.java index c8e735c42158a..fe02975281a46 100644 --- a/solution/0600-0699/0629.K Inverse Pairs Array/Solution.java +++ b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.java @@ -1,22 +1,19 @@ class Solution { - - private static final int MOD = 1000000007; - public int kInversePairs(int n, int k) { - int[] dp = new int[k + 1]; - int[] pre = new int[k + 2]; - for (int i = 1; i <= n; i++) { - dp[0] = 1; - - // dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] - for (int j = 1; j <= k; j++) { - dp[j] = (pre[j + 1] - pre[Math.max(0, j - i + 1)] + MOD) % MOD; + final int mod = (int) 1e9 + 7; + int[] f = new int[k + 1]; + int[] s = new int[k + 2]; + f[0] = 1; + Arrays.fill(s, 1); + s[0] = 0; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= k; ++j) { + f[j] = (s[j + 1] - s[Math.max(0, j - (i - 1))] + mod) % mod; } - - for (int j = 1; j <= k + 1; j++) { - pre[j] = (pre[j - 1] + dp[j - 1]) % MOD; + for (int j = 1; j <= k + 1; ++j) { + s[j] = (s[j - 1] + f[j - 1]) % mod; } } - return dp[k]; + return f[k]; } -} +} \ No newline at end of file diff --git a/solution/0600-0699/0629.K Inverse Pairs Array/Solution.py b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.py index 0215558477a1a..8e62b6c77b7b3 100644 --- a/solution/0600-0699/0629.K Inverse Pairs Array/Solution.py +++ b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.py @@ -1,15 +1,11 @@ class Solution: def kInversePairs(self, n: int, k: int) -> int: - mod = 1000000007 - dp, pre = [0] * (k + 1), [0] * (k + 2) + mod = 10**9 + 7 + f = [1] + [0] * k + s = [0] * (k + 2) for i in range(1, n + 1): - dp[0] = 1 - - # dp[i][j] = dp[i - 1][j - (i - 1)] + ... + dp[i - 1][j] for j in range(1, k + 1): - dp[j] = (pre[j + 1] - pre[max(0, j - i + 1)] + mod) % mod - + f[j] = (s[j + 1] - s[max(0, j - (i - 1))]) % mod for j in range(1, k + 2): - pre[j] = (pre[j - 1] + dp[j - 1]) % mod - - return dp[k] + s[j] = (s[j - 1] + f[j - 1]) % mod + return f[k] diff --git a/solution/0600-0699/0629.K Inverse Pairs Array/Solution.ts b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.ts new file mode 100644 index 0000000000000..b9b11bacc37cb --- /dev/null +++ b/solution/0600-0699/0629.K Inverse Pairs Array/Solution.ts @@ -0,0 +1,16 @@ +function kInversePairs(n: number, k: number): number { + const f: number[] = new Array(k + 1).fill(0); + f[0] = 1; + const s: number[] = new Array(k + 2).fill(1); + s[0] = 0; + const mod: number = 1e9 + 7; + for (let i = 1; i <= n; ++i) { + for (let j = 1; j <= k; ++j) { + f[j] = (s[j + 1] - s[Math.max(0, j - (i - 1))] + mod) % mod; + } + for (let j = 1; j <= k + 1; ++j) { + s[j] = (s[j - 1] + f[j - 1]) % mod; + } + } + return f[k]; +} diff --git a/solution/1600-1699/1630.Arithmetic Subarrays/README.md b/solution/1600-1699/1630.Arithmetic Subarrays/README.md index 86f6d6f621302..93c9d2b17b391 100644 --- a/solution/1600-1699/1630.Arithmetic Subarrays/README.md +++ b/solution/1600-1699/1630.Arithmetic Subarrays/README.md @@ -64,7 +64,7 @@ 函数 $check(nums, l, r)$ 的实现逻辑如下： - 首先，我们计算子数组的长度 $n = r - l + 1$，并将子数组中的元素放入集合 $s$ 中，方便后续的查找； -- 然后，我们计算子数组中的最小值 $a_1$ 和最大值 $a_n$，如果 $a_n - a_1$ 不能被 $n - 1$ 整除，那么子数组不可能形成等差数列，直接返回 $false$；否则，我们计算等差数列的公差 $d = \frac{a_n - a_1}{n - 1}$； +- 然后，我们获取子数组中的最小值 $a_1$ 和最大值 $a_n$，如果 $a_n - a_1$ 不能被 $n - 1$ 整除，那么子数组不可能形成等差数列，直接返回 $false$；否则，我们计算等差数列的公差 $d = \frac{a_n - a_1}{n - 1}$； - 接下来从 $a_1$ 开始，依次计算等差数列中第 $i$ 项元素，如果第 $i$ 项元素 $a_1 + (i - 1) \times d$ 不在集合 $s$ 中，那么子数组不可能形成等差数列，直接返回 $false$；否则，当我们遍历完所有的元素，说明子数组可以重新排列形成等差数列，返回 $true$。 在主函数中，我们遍历所有的查询，对于每个查询 $l[i]$ 和 $r[i]$，我们调用函数 $check(nums, l[i], r[i])$ 判断子数组是否可以重新排列形成等差数列，将结果存入答案数组中。